Introduction


To successfully analyse and solve the equations of Leaving Cert Applied Maths projectiles, one must be very comfortable with trigonometry.

Projectile trigonometry all takes place in [0^\circ,90^\circ] so we should be able to work exclusively in right-angled-triangles (RATs), however I might revert to the unit circle for proofs (without using the unit circle, the definitions for zero and 90^\circ are found by using continuity).

Recalling that two triangles are similar if they have the same angles, the fundamental principle governing trigonometry might be put something like this:

Similar triangles differ only by a scale factor.

We show this below, but what this means is that the ratio of corresponding sides of similar triangles are the same, and if one of the angles is a right-angle, it means that if you have an angle, say 40^\circ, and calculate the ratio of, say, the length of the opposite to the length of the hypotenuse, that your answer doesn’t depend on how large your triangle is and so it makes sense to talk about this ratio for 40^\circ rather than just a specific triangle:

graph1

These are two similar triangles. The opposite/hypotenuse ratio is the same in both cases.

Suppose the dashed triangle is a k-scaled version of the smaller triangle. Then |A'B'|=k|AB| and |A'C'|=k|AC|. Thus the opposite to hypotenuse ratio for the larger triangle is

\displaystyle \frac{|A'B'|}{|A'C'|}=\frac{k|AB|}{k|AC|}=\frac{|AB|}{|AC|},

which is the same as the corresponding ratio for the smaller triangle.

This allows us to define some special ratios, the so-called trigonometric ratios. If you are studying Leaving Cert Applied Maths you know what these are. You should also be aware of the inverse trigonometric functions. Also you should be able to, given the hypotenuse and angle, find comfortably the other two sides. We should also know that sine is maximised at 90^\circ, where it is equal to one.

In projectiles we use another trigonometric ratio:

\displaystyle \sec(\theta)=\frac{\text{hypotenuse}}{\text{adjacent}}=\frac{1}{\cos\theta}.

Note \cos90^\circ=0, so that \sec90^\circ is not defined. Why? Answer here.

The Pythagoras Identity

For any angle \theta,

\sin^2\theta+\cos^2\theta=1.

Proof: Consider a RAT with an angle of \theta and hypotenuse of one:

graph2

We know that \cos \theta=\frac{|BC|}{1}\Rightarrow |BC|=\cos \theta. Similarly |AB|=\sin\theta. The triangle is a RAT so satisfies Pythagoras Theorem:

|BC|^2+|AB|^2=|AC|^2

\Rightarrow (\cos\theta)^2+(\sin\theta)^2=1^2.

The result follows as soon as one realises that \cos^2\theta is a notation for (\cos\theta)^2 \bullet

As a corollary we have the following.

The Sec-Tan Identity

For \theta\neq 90^\circ,

\sec^2\theta=1+\tan^2\theta.

Proof: Divide the Pythagoras Identity by \cos^2\theta (this rules out \theta=90^\circ):

\displaystyle \frac{1}{\cos^2\theta}=1+\frac{\sin^2\theta}{\cos^\theta}.

Note that by the definition of \sec\theta:

\displaystyle \frac{1}{\cos^2\theta}=\frac{1}{\cos\theta\cdot \cos\theta}=\frac{1}{\cos\theta}\cdot\frac{1}{\cos\theta}=\sec^2\theta,

and

\displaystyle\frac{\sin\theta}{\cos\theta}=\frac{\frac{\text{opp}}{\text{hyp}}}{\frac{\text{opp}}{\text{hyp}}}=\frac{\text{opp}}{\text{adj}}=\tan\theta

\displaystyle \Rightarrow \frac{\sin^2\theta}{\cos^2\theta}=\tan^2\theta \bullet

The other identities that we need are a bit more difficult to derive.

The Fundamental Identity

For A>B,

\cos(A-B)=\cos A\cos B+\sin A\sin B.

The identity is true more generally but this suffices for Projectiles.

Proof: Consider the following.

triangle1

The basic objects here are the RATs \Delta ODP and \Delta OCQ. These triangles both have hypotenuse one, \angle DOP=A and \angle COQ=B. Dropping a perpendicular from P to the line segment [OQ], at E gives a third RAT \Delta OEP with a hypotenuse of one with \angle EOP=A-B.

The fact that the hypotenuses are one gives the following:

|DP|=\sin A, |OD|=\cos A, |CQ|=\sin B, |OC|=\cos B, |EP|=\sin(A-B), and |OE|=\cos(A-B).

Now we calculate the length |PQ|^2 in two different ways. First consider the RAT \Delta FPQ. By subtraction

|FQ|=|DP|-|CQ|=\sin A-\sin B and |FP|=|OC|-|OD|=\cos B-\cos A.

Now using the Pythagoras Theorem:

|PQ|^2=(\sin A-\sin B)^2+(\cos B-\cos A)^2

=\sin^2A-2\sin A\sin B+\sin^2B+\cos^2B-2\cos A\cos B+\cos^2B

= \sin^2A+\cos^2A+\sin^2B+\cos^2B-2\sin A\sin B-2\cos A\cos B

=2-2\sin A\sin B-2\cos A\cos B

by the Pythagoras Identity.

Now looking at the RAT \Delta EPQ. From above we know that |EP|=\sin(A-B) while subtraction shows |EQ|=|OQ|-|OE|=1-\cos(A-B). Using Pythagoras

|PQ|^2=(\sin(A-B))^2+(1-\cos(A-B))^2

=\sin^2(A-B)+1-2\cos(A-B)+\cos^2(A-B)

=\cos^2(A-B)+\sin^2(A-B)+1-2\cos(A-B)

=2-2\cos(A-B),

by the Pythagoras Identity. Now equating the two calculations:

2-2\cos A\cos B-2\sin A\sin B=2-2\cos(A-B),

we quickly see the result \bullet

There is a very obvious relationship between sine and cosine:

triangle2

A very quick inspection shows that:

\sin\theta=\cos(90^\circ-\theta) and \cos\theta=\sin(90^\circ-\theta).

We can use this to generate an expression for \sin(A+B).

Sine Addition Formula

\sin(A+B)=\sin A\cos B+\cos A\sin B

Proof: Using the above, and the fundamental identity:

\sin(A+B)=\cos(90^\circ-(A+B))

=\cos((90^\circ-A)-B)

=\cos(90^\circ-A)\cos B+\sin(90^\circ-A)\sin B

=\sin A\cos B+\cos A\sin B \bullet

Sine Double Angle Formula

\sin(2A)=2\sin A\cos A.

Proof: Apply the above with B=A \bullet

We also need a formula for \cos(A+B). It is easy to derive this using the unit circle (basically how to extend sine and cosine beyond just [0^\circ,90^\circ]), but there is an alternative.

Cosine Addition Formula

\cos(A+B)=\cos A\cos B-\sin A\sin B

Proof: Let \theta=A+B and \alpha=B and apply the Fundamental Identity:

\cos(\theta-\alpha)=\cos\theta\cos \alpha+\sin\theta\sin\alpha

\Rightarrow \cos A=\cos(A+B)\cos B+\sin(A+B)\sin B

=\cos (A+B)\cos B+(\sin A\cos B+\sin B\cos A)\sin B

\Rightarrow \cos(A+B)\cos B=\cos A-\sin A\cos B\sin A-\sin^2B\cos A

=\cos A(1-\sin^2B)-\sin A\sin B\cos B.

Using the Pythagoras Identity (how?)

\cos(A+B)\cos B=\cos A\cos^2B-\sin A\sin B\cos B.

If B\neq 90^\circ (which it won’t be in a RAT with A+B\leq 90^\circ), we can divide both sides by \cos B to yield the result \bullet.

It isn’t hard to show the sine difference formula via

\sin(A-B)=\cos(90^\circ-(A-B))=\cos((90^\circ-A)+B),

but I am not sure this formula is ever required necessarily.

You might, upon using the identities above, end up with negative angles. Assuming that negative angles are of the form -A=0^\circ-A, using the difference formulae, along with the values of sine and cosine at zero, shows that

\sin(-A)=-\sin A and \cos(-A)=\cos(A).

Also messing around with the sum and difference formula gives ways of writing the following products:

2\sin A\sin B, 2\sin A\cos B, and 2\cos A\cos B,

as sums.

This is enough to get one through. All of these formulae are in the Examination Tables. The proofs show you where these identities come from: it is by using them that we start to get used to them and begin to recognise them.

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